Slide 1 Colligative Properties Slide 2 Compounds in Aqueous Solution Dissociation - The separation of ions that occurs when an ionic compound dissolves Precipitation Reactions - A chemical reaction in which one or more of the products formed by the combination of ionic substances is only slightly soluble in the solvent (usually water). Net Ionic Equations Ag + + Cl - ----> AgCl (s) Write only those ions/substances that take part in the reaction Spectator ions - they are present, but not involved in the reaction Dissolved Salts ionic form Insoluble Salts undissociated form Molecular Form molecules Ionic substances ions Weak acids/bases molecular form Strong acids/bases ionized form Slide 3 Rule Rules for Net Ionic Equations 1 HCl, HBr, and HI are strong; all other binary acids are weak 2 Ternary acids: If # oxygens > # hydrogens by 2 or more, they are strong; H 2 SO 4, HClO 4 3 Polyprotic acids; 2nd, 3rd,... ionizations are weak 4 Bases: Group IA and IIA hydroxides are strong (except Be); all others are weak 5 Salts: if soluble write in ionic form 6 Oxides: undissociated form 7 Gases: molecular form
Slide 4 Salts and Solutions Salt - a crystalline compound composed of the negative ion of an acid and the positive ion of a base. This is a "neutralization reaction" Acid + Base ----> Salt + H 2 O HCl + NaOH ----> H 2 O + NaCl Salts are also formed by the reaction of acidic or basic anhydrides with a corresponding base, acid or anhydride: Basic Anhydride + Acid: Acidic Anhydride + Base: Basic Anhydride + Acidic Anhydride Na 2 O + H 2 SO 4 ----> Na 2 SO 4 + H 2 O 2NaOH+ SO 3 ----> Na 2 SO 4 + H 2 O Na 2 O + SO 3 ----> Na 2 SO 4 Slide 5 Nomenclature: Salts and Solutions (cont d) Binary acids produce salts ending in "-ide" Ternary acids produce salts ending in "-ate", "-ous", or "- ite Acidic and Basic Salts, each ion is named separately: NaHC 2 O 4 Sodium Hydrogen Oxalate Pb 2 (OH) 2 CO 3 Lead (II) Hydroxide Carbonate Slide 6 Ionization Ionization is the process in which ions are formed from solute molecules by the action of the solvent Hydronium Ion -the H 3 O + ion. It is formed by the addition of a free proton to a water molecule Strong and Weak Electrolytes A strong electrolyte is a compound of which all or almost all of the dissolved material exists as ions in an aqueous solution. An example of this would be hydrogen chloride, sodium chloride. A weak electrolyte is a compound of which a relatively small amount of the dissolved compound exists as ions in an aqueous solution. Acetic acid is a weak electrolyte. It is also referred to as a weak acid for the same reason.
Slide 7 Colligative Properties Colligative Properties are properties of solutions that are determined by the number of particles in solution rather than the type of particles. vapor pressure lowering freezing point depression boiling point elevation rate of diffusion through a membrane (osmosis) Vapor Pressure Lowering Raoult's Law A nonvolatile solute added to a pure solvent will lower the vapor pressure of the solvent. This is because fewer solvent particles are at the surface VP solution = mf solvent x VP pure solvent VP is proportional to the mole fraction of the solvent Slide 8 Problem 1 Sugar in solution. What is the vapor pressure of water at 70 o C if 1.00 x 10 2 g H 2 O dissolved 2 x 10 2 of sucrose (C 12 H 22 O 11, MW = 342 g/mole). First calculate the mole fraction of the water = moles H 2 O / total number of moles of water and sugar 100 g H 2 O/18 g/mole = 5.56 moles H 2 O 200 g sucrose/342 g/mole = 0.585 moles total number of moles = 6.14 moles mole fraction = m.f. = 5.56 moles H2O/6.14 moles solution = 0.905 Using Raoult's Law VP H2O, 70 o C = 31.2 kpa VP solution = m.f. solvent xvp solvent = 0.905 x 31.2 kpa = 28.3 kpa Slide 9 Problem 2 You have a solution of Mercury (I) chloride, HgCl 2, m.f. = 0.163, temp = 25 o C; VP H2O, 25 o C = 3.2 kpa. Calculate the vapor pressure of the solution: VP solution = m.f. solvent xvp solvent = 0.837 x 3.2 kpa = 2.68 kpa
Slide 10 Boiling Point Elevation ( T bp ) and Freezing Point Depression ( Τ fp ) Due to the reduced vapor pressure of the liquids, the freezing point is depressed and the boiling is increased. Calculation of T bp and T fp 1m = 1 molal solution of sugar = 1 mole sugar / 1 kg H 2 O (i=1) 1m = 1 molal solution NaCl = 2 mole solute / 1 kg H 2 O; Na + and Cl - ( i = 2 ) 1m = 1 molal solution CaCl 2 = 3 mole solute / 1 kg H 2 O; Ca +2, and 2 Cl - ( i = 3 ) K bp = molal boiling point constant = 0.515 o C K fp = molal freezing point constant = 1.853 o C T bp = i m K bp T fp = i m K fp Slide 11 Problem Sugar in water. Calculate the freezing point and boiling point for the solution described. Calculate the molality of a sugar solution consisting of 85.0 g sugar (MW 342 g/mole) dissolved in 392 g water (0.392 kg H 2 O). molality = moles sugar / kg water = (85 g / 342 g/mole) / 0.392 kg water = 0.249 mol/0.392 kg = 0.635 m T bp = i m K bp = 0.635 x 0.515 = 0.327 o C boiling point = 100 + T bp = 100 + 0.327 o C = 100.327 o C freezing point = 0 - T fp = 0-1.18 o C = = -1.18 o C Slide 12 Semipermeable membrane Osmotic Pressure only allows certain types of particles to pass through it. Osmosis the movement of solvent through a membrane from an area of higher solvent concentration to an area of lower solvent concentration. Dynamic Equilibrium is finally reached when the rate of flow from the pure solvent side is balanced by the rate of flow from the solute side due to increased back pressure (osmotic pressure) Osmotic pressure results from two things concentration difference of solvent on two sides of a membrane solute particles cannot pass through the membrane whereas solvent particles can.
Slide 13 Electrolytes and Colligative Properties Due to the reduced vapor pressure of the liquids, the freezing point is depressed and the boiling is increased. This can be seen clearly from the diagram in p. 436. The amount of the vapor pressure reduction is based on the molal concentration of the solute. Pay close attention to the number of particles that are in solution for every particle that dissolves.